Consider a measurement value *y* = 2.70ppm with a *standard
uncertainty *of *u*(*y*) = 0.20ppm. (Its expanded uncertainty = *k* x 0.20ppm = 2 x 0.20ppm = 0.40ppm
where coverage factor *k* = 2 at
95% confidence). It is also given that the single tolerance or specification
upper limit of *Tu* = 3.0ppm.

Assuming the normal probability distribution data and a type I error alpha = 0.05 (5%), we are to make a statement of specification conformity at probability of (1-alpha) or 0.95 (95%).

Our **decision rule** is that
: “Acceptance if the hypothesis **Ho: ***P***( y<
3.0ppm) > 0.95**”is true.

Use Microsoft Excel spreadsheet
function: “= 1-NORM.DIST(2.7,3.0,0.2,TRUE)” to calculate *P*(*y*< 3.0ppm) to get 0.933 or 93.3%. Note that the function
“=NORM.DIST(2.7,3.0,0.2,TRUE)” gives the cumulative area under the curve from
far left to right for a value of 0.067 approximately.

Alternatively, we can also calculate a
normalized *z* -value as
(2.7 – 3.0)/0.2 = – 1.50, and look up the one-tailed normal distribution table
for cumulative probability under the curve with *z* =|1.5| which gives 0.5000 + 0.4332 =
0.9332, as a normal distribution curve is symmetrical in shape. See Appendix A
for the normal distribution cumulative table. In fact, we would get the same
answer if we were to use the Excel function “=1- NORM.DIST(-1.5,0,1,TRUE)” as
well.

Since 93.3% < 95.0%, the H_{o}
is rejected, i.e. the sample result of 2.70ppm can be declared non-compliant
with the specification limit, or put it more *mildly*, “not possible to
state compliance” or “conditional pass” or *some other qualification wordings*!

If, for discussion sake, the measured value was 2.60ppm, instead. Would it be within the upper specification limit of 3.0ppm by the above evaluation?

Indeed, by following the above
reasoning, we would find that the normalized z-value as (2.6-3.0)/0.2 = – 2.0
and the cumulative area under the curve was 0.5000 + 0.4772 = 0.977 which is
larger than 0.950. Therefore, the H_{o}
is ** not** rejected, i.e. the sample or test item is declared in
compliant with the specification limit.

*What is the critical acceptable value Xppm
in order not to get H _{o} rejected? *

The task will be simple if we know how
to find the critical *z* -value in
a normal distribution curve where the area under the curve on the right tail is
0.05 out of 1.00, or 5%, as we have fixed our Type I (alpha) risk as 5%.

Reading from the normal distribution
cumulative table in Appendix A, we note that when *z* = 1.645, the area under the curve is
0.5000 + 0.4500 = 0.9500. Similarly, the
absolute value of Excel function “=NORM.INV(0.05,0,1)” also gives a |*z*|-value 0f 1.645.

The critical acceptable value *X* is then calculated as below:

which gives *X *= 2.67ppm.

The conclusion therefore is that any test value found to be less than or equal to 2.67ppm will be declared as in compliance with the specification of 3.0ppm maximum with 95% confidence (or 5% error risk). Any value found larger than 2.67ppm will be assessed for compliant by considering the higher than 5% risk that the test laboratory is willing to undertake, probably based on some commercial reason. In other words, where a confidence level of less than 95% is acceptable to the laboratory, a compliance statement may be possible. Decision is entirely yours!

**Appendix A**

## Leave a Reply